You mix a 100.0 mL sample of a solution that is 0.0131 M in NiCl2 with a 190.0 m

Question

You mix a 100.0 mL sample of a solution that is 0.0131 Min NiCl2 with a 190.0 mL sample of a solution that is 0.250 M in NH3.

After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108

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c g secure l //session.masteringchemistry.com/myct/itemview?offset next&assignmentProblemID; BOSazai2 next&assignmentProblemID; Signed in as Sierra Ma man's hem 142 Spring 2017 ip For Practice 17.15 Enhanced with Feedback previou For Practice 17.15 Enhanced with Feedback Part A After the solution reaches equilibrium what concentration ofNi (aq) remains? The value of Kr for Ni(NI,) 21 is 2.0 x 108 You may want to reference (Ua pages 821-825 Section 17.8 while completing this problem. Express the concentration to two significant figures and include the appropriate units. You mix a 100,0-mL sample of a solution that is 0.0131 M in Nicle with a 190.0-mL sample of a solution that is 0.250 in NH3 M concentration Value Unit My Answers Give Up Submit

Explanation / Answer

No of moles of NiCl2 added = 0.0131 M x 100 x 10-3
                                         = 1.31 x 10-3 mol

No of moles of NH3 added = 0.250 M x 190.0 x 10-3

                                                                    = 0.0475 mol


I am assuming the formula of the complex ion is: [Ni(NH3)6]+2

NiCl2 + 6NH3 == [Ni(NH3)6]+2 + 2Cl-
Since the Kf is very large, assume all the NiCl2 reacts:

therefore no of moles of NH3 reacted = . 1.31 x 10-3 mol x 6

                                                                   = 7.86 x 10-3

Moles of NH3 that remain = 0.0475– 7.86 x 10-3 = 0.03964

Moles of [Ni(NH3)6]+2 that are formed = 0.00134 moles NiCl2 x 1mole [Ni(NH3)6]+2/ 1 mole NiCl2 = 1.31 x 10-3 moles ]Ni(NH3)6]+2.

Final volume = 100.0 mL + 190.0 mL = 290mL = 0.290 Liters

[NH3] = 0.03964 moles/0.290 liters = 0.1367 M
[Ni(NH3)6]+2 = 1.31 x 10-3 moles/0.290 liters = 4.52 x 10-3 M

[Ni(NH3)6{+2}]/ [Ni+2][NH3]^6 = Kf = 2.0 x 10^8

(4.52 x 10-3)/[Ni+2]( 0.1367)^6 = 2.0 x 10^8

[Ni+2] = (4.52 x 10-3)/( 0.1367)^6(2.0 x 10^8)

[Ni+2] = 3.46 x 10^-6 answer

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